Problem 10 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=e^{\left(4 \sqrt{x}+x^{2}\right)}\end{equation}

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = (2x^{-1/2} + 2x) \cdot e^{(4\sqrt{x} + x^2)}\)

Step 1: Identify the Derivative Rule

Since the function is an exponential function of the form \(y = e^u\), where \(u = 4 \sqrt{x} + x^2\), we will use the chain rule for differentiation. The derivative of \(e^u\) with respect to \(x\) is \(e^u \cdot \frac{du}{dx}\).

Step 2: Differentiate the Inner Function \(u\)

First, identify \(u = 4\sqrt{x} + x^2\). Now differentiate \(u\) with respect to \(x\):\[\frac{du}{dx} = \frac{d}{dx}(4\sqrt{x}) + \frac{d}{dx}(x^2)\]This splits the problem into two parts:- \(\frac{d}{dx}(4\sqrt{x}) = 4 \cdot \frac{d}{dx}(x^{1/2}) = 4 \times \frac{1}{2}x^{-1/2} = 2x^{-1/2}\)- \(\frac{d}{dx}(x^2) = 2x\)Thus, \(\frac{du}{dx} = 2x^{-1/2} + 2x\).

Step 3: Apply Chain Rule to Find \(\frac{dy}{dx}\)

Now apply the chain rule:\[\frac{dy}{dx} = e^{(4\sqrt{x} + x^2)} \cdot (2x^{-1/2} + 2x)\]This multiplying results in:\[\frac{dy}{dx} = e^{(4\sqrt{x} + x^2)} \cdot (2x^{-1/2} + 2x)\]

Step 4: Simplify Expression

The expression from the previous step can be simplified to:\[\frac{dy}{dx} = (2x^{-1/2} + 2x) \cdot e^{(4\sqrt{x} + x^2)}\]

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus for finding the derivative of composite functions. In simple terms, it helps us understand how the rate of change of one function affects another function that depends on it.
For a function that can be expressed as the composition of two or more functions, such as \( y = e^{u(x)} \), the chain rule comes into play. Here's how it works:

• Identify the outside function and the inside function. In this exercise, the outside function is the exponential function \( e^u \), and the inside function is \( u = 4\sqrt{x} + x^2 \).
• First, differentiate the outside function with respect to its own variable, keeping the inside function unchanged, which gives \( e^u \).
• Next, differentiate the inside function. In this case, \( \frac{du}{dx} = 2x^{-1/2} + 2x \).
• Finally, multiply these derivatives together to obtain the derivative of the entire expression: \( \frac{dy}{dx} = e^u \cdot \frac{du}{dx} \).

The chain rule is crucial in differentiating such compositions of functions, making it a powerful tool in calculus.

Exponential Function

Exponential functions are a core topic in calculus. They are expressed in the form \( y = b^x \), where \( b \) is a positive real number, and \( x \) is a variable. A special case in calculus is the natural exponential function, \( y = e^x \), where \( e \approx 2.718 \) is the base of the natural logarithm.
This special function has important properties that make it straightforward to differentiate. Specifically, the derivative of \( e^x \) is simply \( e^x \). That means, if you have a function \( y = e^{f(x)} \), the differentiation revolves around the use of the chain rule, taking into account the inner derivative \( f'(x) \).

• For example, in the given solution, our function is \( y = e^{(4\sqrt{x} + x^2)} \).
• Here, \( 4\sqrt{x} + x^2 \) is the exponent, and the exponential rule combined with the chain rule aids in differentiating it smoothly.

Understanding exponential functions and their derivatives is vital for solving a wide array of calculus problems.

Differentiation

Differentiation is a fundamental concept in calculus, used to determine the rate of change of a function regarding its variable. It is the process of finding the derivative, denoted as \( \frac{dy}{dx} \). This helps us understand how a function behaves and changes.
The basic steps of differentiation involve:

• Identifying the type of function and selecting the proper differentiation rule (e.g., power rule, product rule, chain rule).
• Applying the rule systematically to find the derivative.

When a function involves more complex compositions, like combinations or transformations of standard functions, differentiation might not be straightforward.

• For instance, in our exercise, to find \( \frac{du}{dx} \) for \( u = 4\sqrt{x} + x^2 \), we used differentiation rules for both square roots and power functions: \( \frac{d}{dx}(x^{1/2}) \) and \( \frac{d}{dx}(x^2) \).

Mastering differentiation rules and techniques is crucial for solving calculus problems efficiently.

Calculus Problem Solving

Problem-solving in calculus often involves strategically breaking down the problem into manageable steps, especially when dealing with differentiating functions involving multiple components. In our exercise, each step was carefully executed to avoid errors and provide clarity.
Here’s a typical approach:

• Identify and understand the problem. Recognize the components of the function that determine which rules to apply.
• Plan your solution path by listing which rules are necessary, like the chain rule for compositions, and ensuring the calculation order is clear.
• Execute step-by-step calculations while double-checking each step to avoid mistakes.
• Finally, review the resulting expressions, simplifying where possible to ensure the answer accurately represents the derivative of the original function.

Developing a structured problem-solving strategy helps in tackling more complex calculus problems successfully, leveraging fundamental concepts like the chain rule and exponential function differentiation.